从「越来越接近」到定义

上一讲把 $\sqrt 2$ 写成截断序列的极限，但「极限」还只是直觉。本讲把它变成可以检验的定义。

直觉版本是：$a_n$ 越来越接近 $a$。问题是「越来越」没有量化——接近多少算接近？前一百项很乱算不算数？

定义的做法是：把「接近」交给挑战者规定。对方给出任意精度 $\varepsilon>0$，你只要能找到一个门槛 $N$，使得门槛之后的所有项都落在 $a$ 的 $\varepsilon$ 邻域里。能应付任意 $\varepsilon$，就叫收敛。

时间轴

收敛的定义

定义。 数列 $(a_n)$ 收敛到 $a$，如果

$$\forall \varepsilon>0\ \ \exists N\in\mathbb N\ \ \forall n\ge N:\quad |a_n-a|<\varepsilon .$$

此时记 $a_n\to a$ 或 $\lim_{n\to\infty}a_n=a$。不收敛的数列称为发散。

读法很关键：前两个量词属于不同的人。$\varepsilon$ 由对方给出，$N$ 由你给出；$N$ 可以依赖 $\varepsilon$，但不能依赖 $n$。一旦 $N$ 选定，门槛之后的所有 $n$ 都必须满足要求。这就是「最终」的意思：有限个例外不影响收敛。

几何图像：在实轴上以 $a$ 为中心、$\varepsilon$ 为半径画区间 $(a-\varepsilon,a+\varepsilon)$。定义说，无论区间多窄，数列最终都会整个掉进去，并且再也不出来。

三个例子

$a_n=(-1)^n$ 发散

这个数列在 $-1$ 与 $1$ 之间来回跳。它不收敛，因为找不到一个 $a$ 能让所有足够靠后的项都靠近它：无论 $a$ 取什么，$\varepsilon=1/2$ 的邻域都装不下 $-1$ 和 $1$ 两个值，而这两类项在任意长的尾巴里都反复出现。

严格写法：若 $a_n\to a$，取 $\varepsilon=1/2$，存在 $N$。取偶数 $n\ge N$，则 $|1-a|<1/2$；取奇数 $m\ge N$，则 $|-1-a|<1/2$。于是

$$2=|1-(-1)|\le|1-a|+|a-(-1)|<\frac12+\frac12=1,$$

矛盾。

$a_n=1/n\to0$

给定 $\varepsilon>0$，由阿基米德性质取 $N\in\mathbb N$ 使 $N>1/\varepsilon$，即 $1/N<\varepsilon$。则当 $n\ge N$ 时，

$$\left|\frac1n-0\right|=\frac1n\le\frac1N<\varepsilon .$$

所以 $1/n\to0$。这个例子展示了标准套路：先用阿基米德性质把 $N$ 造出来，再验证。

$0.999\ldots=1$

把无限小数定义为截断序列

$$a_n=\underbrace{0.999\ldots9}_{n\text{ 个 }9}=1-10^{-n}.$$

要证 $a_n\to1$。给定 $\varepsilon>0$，取 $N$ 使 $1/N<\varepsilon$。当 $n\ge N$ 时，

$$|a_n-1|=10^{-n}\le\frac1n\le\frac1N<\varepsilon .$$

（第一个不等号因为 $10^n\ge n$，可对 $n$ 归纳验证。）所以 $0.999\ldots=1$。

这里有一个值得注意的地方：每个截断都严格小于 $1$，但极限等于 $1$。极限不必等于序列中的任何一项，它只描述尾巴的整体行为。

收敛必有界

定理。 若 $(a_n)$ 收敛，则集合 $\{a_n:n\in\mathbb N\}$ 有界。

证明。 设 $a_n\to a$。取 $\varepsilon=1$，存在 $N_0$，使 $n\ge N_0$ 时 $|a_n-a|<1$。由三角不等式，

$$|a_n|=|a_n-a+a|\le|a_n-a|+|a|<1+|a| \quad (n\ge N_0).$$

剩下的只有前 $N_0-1$ 项，是有限集。令

$$C=\max\{|a_1|,|a_2|,\dots,|a_{N_0-1}|\}$$

（若 $N_0=1$，取 $C=0$）。则对一切 $n$，

$$|a_n|\le\max\{C,\,|a|+1\},$$

故数列有界。$\square$

证明的结构值得记住：尾巴无限但整齐，头部有限但乱。用一个固定的 $\varepsilon$ 把尾巴压住，再对有限的头部取最大值。后面许多证明都沿用这个「切成两段」的思路。

逆命题不成立：$(-1)^n$ 有界但发散。所以有界只是收敛的必要条件，不能用来证明收敛。

极限的四则运算

设 $a_n\to a$、$b_n\to b$。则

1. $ca_n\to ca$（$c\in\mathbb R$）；
2. $a_n+b_n\to a+b$；
3. $a_nb_n\to ab$；
4. 若所有 $a_n\ne0$ 且 $a\ne0$，则 $1/a_n\to1/a$。

课程把 (1)(2) 写成正式证明，(3)(4) 给详细思路。

(1) 数乘

$c=0$ 时两边都是 $0$，平凡。设 $c\ne0$。给定 $\varepsilon>0$，则 $\varepsilon/|c|>0$。由 $a_n\to a$，存在 $N$，使 $n\ge N$ 时 $|a_n-a|<\varepsilon/|c|$。两边乘以 $|c|$：

$$|ca_n-ca|=|c|\,|a_n-a|<\varepsilon .$$

所以 $ca_n\to ca$。

关键动作：把对方给的 $\varepsilon$ 换算成自己需要的精度 $\varepsilon/|c|$。定义里的 $\varepsilon$ 是任意的，所以可以代换成任何正数。

(2) 加法

给定 $\varepsilon>0$。由 $a_n\to a$，存在 $N_a$，使 $n\ge N_a$ 时 $|a_n-a|<\varepsilon/2$；由 $b_n\to b$，存在 $N_b$，使 $n\ge N_b$ 时 $|b_n-b|<\varepsilon/2$。

取 $N=\max\{N_a,N_b\}$。当 $n\ge N$ 时两个不等式同时成立，于是

$$|(a_n+b_n)-(a+b)|\le|a_n-a|+|b_n-b|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon .$$

所以 $a_n+b_n\to a+b$。

要点：两个误差各分一半，门槛取最大。门槛取最大是因为要同时满足两个条件。

(3) 乘法：插入中间项

直接比较 $a_nb_n$ 与 $ab$ 很困难，因为两边都在动。技巧是插入 $ab_n$，把它拆成两段：

$$a_nb_n-ab=a_nb_n-ab_n+ab_n-ab=a_n(b_n-b)+b(a_n-a).$$

这样每一段都只有一个「动」的因子：

$$|a_nb_n-ab|\le|a_n|\,|b_n-b|+|b|\,|a_n-a| .$$

第二段乘以固定常数 $|b|$，容易控制。第一段的 $|a_n|$ 不是常数，但上一节证明了收敛数列有界：存在 $C$，使 $|a_n|\le C$ 对一切 $n$ 成立。于是

$$|a_nb_n-ab|\le C|b_n-b|+|b|\,|a_n-a| .$$

现在给定 $\varepsilon>0$。取

$$|b_n-b|<\frac{\varepsilon}{2(C+1)}, \qquad |a_n-a|<\frac{\varepsilon}{2(|b|+1)}$$

分别在 $n\ge N_b$、$n\ge N_a$ 时成立，再令 $N=\max\{N_a,N_b\}$。当 $n\ge N$ 时，

$$|a_nb_n-ab|<\frac{C\varepsilon}{2(C+1)}+\frac{|b|\varepsilon}{2(|b|+1)}<\frac\varepsilon2+\frac\varepsilon2=\varepsilon .$$

所以 $a_nb_n\to ab$。

分母写成 $C+1$ 与 $|b|+1$ 而不是 $C$、$|b|$，是为了避免除以零：即使 $C=0$ 或 $b=0$，式子仍然有意义。

这个插入技巧会反复出现。 课程提到，证明乘积求导法则（Leibniz 法则）时用的是同一个拆分：$f(x+h)g(x+h)-f(x)g(x)$ 中间插入 $f(x+h)g(x)$。

(4) 倒数：分母远离零

要证 $1/a_n\to1/a$。先通分：

$$\left|\frac1{a_n}-\frac1a\right| =\left|\frac{a-a_n}{a_na}\right| =\frac{|a_n-a|}{|a_n|\,|a|}.$$

分子已经会趋于 $0$，麻烦在分母：$|a_n|$ 可能很小，把商放大。所以要证明充分靠后时 $|a_n|$ 不会太小。

由 $a\ne0$，有 $|a|/2>0$。因为 $a_n\to a$，存在 $N_1$，使 $n\ge N_1$ 时

$$|a_n-a|<\frac{|a|}{2}.$$

由反三角不等式 $|a|-|a_n|\le|a-a_n|$，得

$$|a_n|>|a|-\frac{|a|}{2}=\frac{|a|}{2}\quad (n\ge N_1).$$

于是当 $n\ge N_1$ 时，

$$\left|\frac1{a_n}-\frac1a\right| =\frac{|a_n-a|}{|a_n|\,|a|} <\frac{|a_n-a|}{(|a|/2)\cdot|a|} =\frac{2|a_n-a|}{|a|^2}.$$

最后给定 $\varepsilon>0$，取 $N\ge N_1$ 使 $n\ge N$ 时 $|a_n-a|<\dfrac{\varepsilon|a|^2}{2}$，就得到

$$\left|\frac1{a_n}-\frac1a\right|<\varepsilon .$$

所以 $1/a_n\to1/a$。

核心只有一句：极限非零，所以数列最终被推离零至少 $|a|/2$。这是所有「除法」论证的共同前提。

子列

定义

设 $(a_n)$ 是数列，$g:\mathbb N\to\mathbb N$ 严格递增。则

$$b_k=a_{g(k)}$$

称为 $(a_n)$ 的子列，常记作 $(a_{n_k})$。

严格递增意味着 $n_1<n_2<n_3<\cdots$：只能向后走，不能回头，也不能原地停。这正好排除两类「假子列」：

• 重复取同一项，例如从 $1,2,3,4,\dots$ 中取 $1,1,2,\dots$；
• 打乱顺序，例如取 $3,2,4,5,\dots$。

课程给的合法例子：从 $(-1)^n$ 中取全部偶数项，得到常数列 $1,1,1,\dots$；取全部奇数项，得到 $-1,-1,-1,\dots$；也可以取「前两个奇数、再两个偶数」这种混合模式，只要指标严格递增。

定理：收敛当且仅当所有子列收敛

定理。 $a_n\to a$ 当且仅当 $(a_n)$ 的每个子列都收敛到 $a$。

证明。 「$\Leftarrow$」是平凡的：取 $g(k)=k$（严格递增），对应的子列就是原数列本身。若所有子列都收敛到 $a$，原数列自然也收敛到 $a$。

「$\Rightarrow$」：设 $a_n\to a$，$(a_{n_k})$ 是任一子列。给定 $\varepsilon>0$，存在 $N$，使 $n\ge N$ 时 $|a_n-a|<\varepsilon$。

因为 $g$ 严格递增且取自然数值，可以归纳证明 $n_k\ge k$：$n_1\ge1$，且每步至少增加 $1$。所以当 $k\ge N$ 时，$n_k\ge k\ge N$，从而

$$|a_{n_k}-a|<\varepsilon .$$

故 $a_{n_k}\to a$。$\square$

这条定理最常用的形式是它的逆否命题：若能找到两个收敛到不同极限的子列，原数列一定发散。例如 $(-1)^n$ 的偶数项收敛到 $1$、奇数项收敛到 $-1$，所以它发散——这比直接验证定义省事得多。

要注意的是，只有一个收敛子列不能推出原数列收敛。常数子列可以从任何数列里挑出来（只要某个值出现无穷多次），但原数列可能仍然乱跳。

本节结论

• $a_n\to a$ 定义为：$\forall\varepsilon>0\ \exists N\ \forall n\ge N$，$|a_n-a|<\varepsilon$；$N$ 可以依赖 $\varepsilon$，不能依赖 $n$。
• 收敛只约束尾巴，有限个例外无关紧要；极限不必是数列中的某一项。
• 收敛必有界；有界不一定收敛。
• 四则运算：数乘换算 $\varepsilon/|c|$；加法各分 $\varepsilon/2$ 并取最大门槛；乘法插入 $ab_n$ 并用有界性；倒数依靠 $|a_n|>|a|/2$。
• 子列由严格递增的指标映射给出；子列保留顺序、不能重复。
• $a_n\to a$ 等价于所有子列都收敛到 $a$；两个子列极限不同即可判定发散。

理解检验

1. 为什么定义里的 $N$ 可以依赖 $\varepsilon$，却不能依赖 $n$？如果允许 $N$ 依赖 $n$，定义会变成什么样？

参考答案：收敛要求「存在一个门槛，从它之后所有项都满足要求」。$N$ 依赖 $\varepsilon$ 是允许的，因为对方先给出精度，你据此决定门槛；$N$ 不能依赖 $n$，因为必须一次选定、对所有更大的 $n$ 同时有效。若允许 $N=N(n)$，则对每个 $n$ 都能取 $N=n$，条件退化为「$|a_n-a|<\varepsilon$」，任何数列都满足——定义将失去内容。

2. 用定义证明 $a_n=\dfrac{2n+1}{n}\to2$。

参考答案：先化简 $|a_n-2|=\left|\frac{2n+1-2n}{n}\right|=\frac1n$。给定 $\varepsilon>0$，由阿基米德性质取 $N\in\mathbb N$ 使 $N>1/\varepsilon$。当 $n\ge N$ 时，$|a_n-2|=\frac1n\le\frac1N<\varepsilon$。故 $a_n\to2$。

3. 证明收敛必有界时，为什么必须把数列切成「前 $N_0-1$ 项」和「尾巴」两段？只用 $\varepsilon=1$ 会漏掉什么？

参考答案：$\varepsilon=1$ 只对 $n\ge N_0$ 的项给出 $|a_n|<|a|+1$，这条界管不到前面的有限项——它们可能非常大，且与 $a$ 无关。因此必须单独处理：前 $N_0-1$ 项只有有限多个，取其中绝对值最大者 $C$；再把尾巴的界 $|a|+1$ 与 $C$ 取最大，得到覆盖全体的界。漏掉这一步，得到的只是一个「最终有界」的结论，而不是整条数列有界。

4. 证明乘积极限时，为什么要插入 $ab_n$？为什么必须先证明 $(a_n)$ 有界？

参考答案：直接估计 $|a_nb_n-ab|$ 时两个因子都在变化，无法分离。插入 $ab_n$ 后拆成 $a_n(b_n-b)+b(a_n-a)$：第一段只有一个变化的因子 $b_n-b$，第二段只有一个变化的因子 $a_n-a$，且 $b$ 是常数。估计第一段时需要 $|a_n|\le C$ 把变化的系数换成常数；有界性正是收敛数列提供的。没有它，第一段的界会随 $n$ 变化，无法统一小于 $\varepsilon/2$。

5. 若一个数列有两个子列分别收敛到不同的极限，原数列是否收敛？反过来，若存在一个收敛子列，能否推出原数列收敛？

参考答案：第一种情况原数列一定发散。因为若 $a_n\to a$，由定理每个子列都收敛到 $a$，不可能出现两个不同极限。第二种情况不能推出收敛：$(-1)^n$ 的偶数项子列恒为 $1$，收敛，但原数列发散。存在一个收敛子列只说明数列在某个「稀疏」的方向上有规律，不能约束其余项。

6. 用子列判定 $a_n=(-1)^n+\frac1n$ 发散。

参考答案：取偶数项子列 $a_{2k}=1+\frac1{2k}\to1$，奇数项子列 $a_{2k-1}=-1+\frac1{2k-1}\to-1$。两个子列都收敛但极限不同，由定理的逆否命题，原数列发散。这与直接使用 $\varepsilon$ 定义相比更简洁，因为不必对任意候选极限逐一否定。

与前后课的关系

本讲给出了收敛的定义与基本工具。下一讲用完备性证明单调收敛定理：单调有界数列必收敛——它第一次把「有界」这个较弱的条件变成收敛的充分条件。第 6 讲的柯西准则则把「极限存在」改写成不依赖极限值的条件，为第 7 讲的 Bolzano–Weierstrass 定理铺路。

一条主线逐渐清晰：定义（本讲）→ 充分条件（单调 + 有界）→ 内蕴判据（柯西）→ 紧性（Bolzano–Weierstrass）。

From "getting closer" to a definition

The previous lecture wrote $\sqrt2$ as the limit of its truncations, but "limit" was still intuition. This lecture turns it into something checkable.

The intuitive version says: $a_n$ gets closer and closer to $a$. The trouble is that "closer" is not quantified — how close counts, and do the first hundred messy terms matter?

The definition hands the notion of closeness to a challenger. Whatever accuracy $\varepsilon>0$ is demanded, you must produce a threshold $N$ beyond which every term lies within $\varepsilon$ of $a$. Surviving every $\varepsilon$ is what convergence means.

Timeline

The definition

Definition. A sequence $(a_n)$ converges to $a$ if

$$\forall \varepsilon>0\ \ \exists N\in\mathbb N\ \ \forall n\ge N:\quad |a_n-a|<\varepsilon .$$

One writes $a_n\to a$ or $\lim_{n\to\infty}a_n=a$. A sequence that does not converge diverges.

The reading matters: the first two quantifiers belong to different people. $\varepsilon$ is chosen by the challenger, $N$ by you; $N$ may depend on $\varepsilon$ but not on $n$. Once $N$ is fixed, every later index must satisfy the demand. That is what "eventually" means: finitely many exceptions are irrelevant.

Geometrically, draw the interval $(a-\varepsilon,a+\varepsilon)$ around $a$. The definition says that however narrow the interval, the sequence eventually falls inside it and stays there.

Three examples

$a_n=(-1)^n$ diverges

The sequence jumps between $-1$ and $1$. No candidate $a$ can capture it: for $\varepsilon=1/2$, the interval around $a$ cannot contain both $-1$ and $1$, and both values recur in every long tail.

Formally: suppose $a_n\to a$ and take $\varepsilon=1/2$, giving $N$. For even $n\ge N$, $|1-a|<1/2$; for odd $m\ge N$, $|-1-a|<1/2$. Then

$$2=|1-(-1)|\le|1-a|+|a-(-1)|<\frac12+\frac12=1,$$

a contradiction.

$a_n=1/n\to0$

Given $\varepsilon>0$, the Archimedean property provides $N\in\mathbb N$ with $N>1/\varepsilon$, i.e. $1/N<\varepsilon$. For $n\ge N$,

$$\left|\frac1n-0\right|=\frac1n\le\frac1N<\varepsilon .$$

So $1/n\to0$. This is the standard pattern: build $N$ from the Archimedean property, then verify.

$0.999\ldots=1$

Define the infinite decimal through its truncations

$$a_n=\underbrace{0.999\ldots9}_{n\text{ nines}}=1-10^{-n}.$$

Given $\varepsilon>0$, choose $N$ with $1/N<\varepsilon$. For $n\ge N$,

$$|a_n-1|=10^{-n}\le\frac1n\le\frac1N<\varepsilon .$$

(The first inequality holds because $10^n\ge n$, provable by induction.) Hence $0.999\ldots=1$.

Note something important: every truncation is strictly below $1$, yet the limit equals $1$. A limit need not be one of the terms; it describes only the behaviour of the tail.

Convergent sequences are bounded

Theorem. If $(a_n)$ converges, then the set $\{a_n:n\in\mathbb N\}$ is bounded.

Proof. Let $a_n\to a$. Take $\varepsilon=1$; there is $N_0$ with $|a_n-a|<1$ for $n\ge N_0$. By the triangle inequality,

$$|a_n|=|a_n-a+a|\le|a_n-a|+|a|<1+|a| \quad (n\ge N_0).$$

Only the first $N_0-1$ terms remain, a finite set. Let

$$C=\max\{|a_1|,|a_2|,\dots,|a_{N_0-1}|\}$$

(with $C=0$ if $N_0=1$). Then for every $n$,

$$|a_n|\le\max\{C,\,|a|+1\},$$

so the sequence is bounded. $\square$

The structure is worth remembering: the tail is infinite but orderly, the head finite but wild. One fixed $\varepsilon$ controls the tail; a maximum controls the head. Many later proofs reuse this cut-in-two device.

The converse fails: $(-1)^n$ is bounded yet divergent. Boundedness is necessary, not sufficient.

The limit laws

Suppose $a_n\to a$ and $b_n\to b$. Then

1. $ca_n\to ca$ for $c\in\mathbb R$;
2. $a_n+b_n\to a+b$;
3. $a_nb_n\to ab$;
4. if all $a_n\ne0$ and $a\ne0$, then $1/a_n\to1/a$.

The lecture writes (1) and (2) as formal proofs and gives detailed outlines for (3) and (4).

(1) Scalar multiples

If $c=0$ both sides vanish and the claim is trivial. Let $c\ne0$. Given $\varepsilon>0$, the number $\varepsilon/|c|$ is positive. Since $a_n\to a$, there is $N$ with $|a_n-a|<\varepsilon/|c|$ for $n\ge N$. Multiplying by $|c|$,

$$|ca_n-ca|=|c|\,|a_n-a|<\varepsilon .$$

Hence $ca_n\to ca$.

The key move: convert the given $\varepsilon$ into the accuracy you need, $\varepsilon/|c|$. Since the definition holds for every positive number, any positive substitute is allowed.

(2) Sums

Given $\varepsilon>0$. Since $a_n\to a$ there is $N_a$ with $|a_n-a|<\varepsilon/2$ for $n\ge N_a$; since $b_n\to b$ there is $N_b$ with $|b_n-b|<\varepsilon/2$ for $n\ge N_b$.

Set $N=\max\{N_a,N_b\}$. For $n\ge N$ both inequalities hold, so

$$|(a_n+b_n)-(a+b)|\le|a_n-a|+|b_n-b|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon .$$

Hence $a_n+b_n\to a+b$.

Two errors are split in half, and the thresholds are combined by taking the maximum — because both conditions must hold simultaneously.

(3) Products: insert a middle term

Comparing $a_nb_n$ with $ab$ directly is hard because both factors move. The trick is to insert $ab_n$:

$$a_nb_n-ab=a_nb_n-ab_n+ab_n-ab=a_n(b_n-b)+b(a_n-a).$$

Each piece now has only one moving factor:

$$|a_nb_n-ab|\le|a_n|\,|b_n-b|+|b|\,|a_n-a| .$$

The second piece carries the constant $|b|$ and is easy. The first carries $|a_n|$, which is not constant, but the previous theorem gives boundedness: there is $C$ with $|a_n|\le C$ for all $n$. Hence

$$|a_nb_n-ab|\le C|b_n-b|+|b|\,|a_n-a| .$$

Now given $\varepsilon>0$, arrange

$$|b_n-b|<\frac{\varepsilon}{2(C+1)}, \qquad |a_n-a|<\frac{\varepsilon}{2(|b|+1)}$$

for $n\ge N_b$ and $n\ge N_a$ respectively, and set $N=\max\{N_a,N_b\}$. For $n\ge N$,

$$|a_nb_n-ab|<\frac{C\varepsilon}{2(C+1)}+\frac{|b|\varepsilon}{2(|b|+1)}<\frac\varepsilon2+\frac\varepsilon2=\varepsilon .$$

So $a_nb_n\to ab$.

The denominators $C+1$ and $|b|+1$ rather than $C$ and $|b|$ avoid division by zero: the expressions stay meaningful even when $C=0$ or $b=0$.

This insertion trick recurs. The lecture notes that the product rule for derivatives (Leibniz) uses the same split, inserting $f(x+h)g(x)$ into $f(x+h)g(x+h)-f(x)g(x)$.

(4) Reciprocals: the denominator stays away from zero

We want $1/a_n\to1/a$. First put over a common denominator:

$$\left|\frac1{a_n}-\frac1a\right| =\left|\frac{a-a_n}{a_na}\right| =\frac{|a_n-a|}{|a_n|\,|a|}.$$

The numerator tends to $0$; the danger is the denominator, since a small $|a_n|$ inflates the quotient. So we must show that eventually $|a_n|$ is not small.

Since $a\ne0$, $|a|/2>0$. Because $a_n\to a$, there is $N_1$ with

$$|a_n-a|<\frac{|a|}{2}\quad (n\ge N_1).$$

By the reverse triangle inequality $|a|-|a_n|\le|a-a_n|$,

$$|a_n|>|a|-\frac{|a|}{2}=\frac{|a|}{2}\quad (n\ge N_1).$$

Thus for $n\ge N_1$,

$$\left|\frac1{a_n}-\frac1a\right| =\frac{|a_n-a|}{|a_n|\,|a|} <\frac{|a_n-a|}{(|a|/2)\cdot|a|} =\frac{2|a_n-a|}{|a|^2}.$$

Finally, given $\varepsilon>0$, choose $N\ge N_1$ so that $|a_n-a|<\dfrac{\varepsilon|a|^2}{2}$ for $n\ge N$. Then

$$\left|\frac1{a_n}-\frac1a\right|<\varepsilon .$$

Hence $1/a_n\to1/a$.

The core fact is one sentence: a nonzero limit eventually pushes the sequence at least $|a|/2$ away from zero. Every division argument rests on it.

Subsequences

Definition

Let $(a_n)$ be a sequence and $g:\mathbb N\to\mathbb N$ strictly increasing. Then

$$b_k=a_{g(k)}$$

is a subsequence of $(a_n)$, usually written $(a_{n_k})$.

Strict increase means $n_1<n_2<n_3<\cdots$: only forward, never backward, never standing still. This rules out two fake subsequences:

• repeating a term, e.g. taking $1,1,2,\dots$ from $1,2,3,4,\dots$;
• reordering, e.g. taking $3,2,4,5,\dots$.

Legal examples from $(-1)^n$: all even indices give the constant sequence $1,1,1,\dots$; all odd indices give $-1,-1,-1,\dots$; a mixed pattern such as "two odd, then two even" is also legal, provided the indices strictly increase.

Theorem: convergence iff every subsequence converges

Theorem. $a_n\to a$ if and only if every subsequence of $(a_n)$ converges to $a$.

Proof. "$\Leftarrow$" is trivial: take $g(k)=k$ (strictly increasing); the corresponding subsequence is the original sequence. If every subsequence converges to $a$, the sequence itself does.

"$\Rightarrow$": let $a_n\to a$ and let $(a_{n_k})$ be any subsequence. Given $\varepsilon>0$, there is $N$ with $|a_n-a|<\varepsilon$ for $n\ge N$.

Because $g$ is strictly increasing with natural values, induction gives $n_k\ge k$: $n_1\ge1$ and each step increases by at least $1$. Hence for $k\ge N$, $n_k\ge k\ge N$, so

$$|a_{n_k}-a|<\varepsilon .$$

Therefore $a_{n_k}\to a$. $\square$

The most useful form is the contrapositive: two subsequences with different limits force divergence. For $(-1)^n$, the even terms converge to $1$ and the odd terms to $-1$, so the sequence diverges — far less work than negating the definition directly.

Note that a single convergent subsequence does not imply convergence of the original. A constant subsequence can be extracted whenever some value occurs infinitely often, while the original may still jump around.

What this lecture establishes

• $a_n\to a$ means: $\forall\varepsilon>0\ \exists N\ \forall n\ge N$, $|a_n-a|<\varepsilon$; $N$ may depend on $\varepsilon$, not on $n$.
• Convergence constrains only the tail; finitely many exceptions are irrelevant, and the limit need not be a term.
• Convergent implies bounded; bounded does not imply convergent.
• Limit laws: scale $\varepsilon$ by $|c|$; split $\varepsilon/2$ and take the maximum threshold for sums; insert $ab_n$ and use boundedness for products; use $|a_n|>|a|/2$ for reciprocals.
• A subsequence comes from a strictly increasing index map; it preserves order and never repeats.
• Convergence is equivalent to every subsequence converging to the same limit; two different subsequential limits prove divergence.

Comprehension check

1. Why may $N$ depend on $\varepsilon$ but not on $n$? What would the definition become if $N$ could depend on $n$?

Reference answer: Convergence requires that one threshold works for every sufficiently large index. Dependence on $\varepsilon$ is allowed because the challenger fixes the accuracy first and you respond with a threshold. Dependence on $n$ would let you choose $N=n$ for each term, reducing the condition to $|a_n-a|<\varepsilon$ for every $n$ — which every sequence satisfies, so the definition would carry no content.

2. Prove from the definition that $a_n=\dfrac{2n+1}{n}\to2$.

Reference answer: Simplify first: $|a_n-2|=\left|\frac{2n+1-2n}{n}\right|=\frac1n$. Given $\varepsilon>0$, the Archimedean property gives $N\in\mathbb N$ with $N>1/\varepsilon$. For $n\ge N$, $|a_n-2|=\frac1n\le\frac1N<\varepsilon$. Hence $a_n\to2$.

3. Why must the boundedness proof split the sequence into the first $N_0-1$ terms and the tail? What does using only $\varepsilon=1$ miss?

Reference answer: Taking $\varepsilon=1$ yields $|a_n|<|a|+1$ only for $n\ge N_0$; it says nothing about the finitely many earlier terms, which may be arbitrarily large and unrelated to $a$. Those must be handled separately: they form a finite set, so take $C$ to be their maximum absolute value, then take the maximum of $C$ and $|a|+1$ to cover everything. Without this step one only proves eventual boundedness, not boundedness of the whole sequence.

4. In the product proof, why insert $ab_n$, and why prove boundedness of $(a_n)$ first?

Reference answer: Directly estimating $|a_nb_n-ab|$ leaves two varying factors that cannot be separated. Inserting $ab_n$ splits it into $a_n(b_n-b)+b(a_n-a)$: the first piece has only $b_n-b$ varying, the second only $a_n-a$ varying, with $b$ constant. Controlling the first piece requires replacing $|a_n|$ by a constant $C$, which is exactly what boundedness supplies. Without it the bound for the first piece would itself vary with $n$ and could not be made uniformly less than $\varepsilon/2$.

5. If two subsequences converge to different limits, does the original sequence converge? Conversely, does one convergent subsequence imply convergence?

Reference answer: In the first case the original diverges: if $a_n\to a$, the theorem makes every subsequence converge to $a$, so two distinct limits are impossible. In the second case convergence does not follow: the even-indexed subsequence of $(-1)^n$ is constantly $1$ and converges, yet the sequence diverges. One convergent subsequence only shows regularity along a sparse set of indices.

6. Use subsequences to show that $a_n=(-1)^n+\frac1n$ diverges.

Reference answer: The even terms give $a_{2k}=1+\frac1{2k}\to1$; the odd terms give $a_{2k-1}=-1+\frac1{2k-1}\to-1$. Both subsequences converge, to different limits, so by the contrapositive the sequence diverges. This is simpler than negating the $\varepsilon$ definition for every candidate limit.

Relation to neighbouring lectures

This lecture supplies the definition and the basic toolkit. The next lecture uses completeness to prove the monotone convergence theorem: a monotone bounded sequence converges — the first time "bounded" becomes sufficient. Lecture 6's Cauchy criterion rewrites existence of a limit without referring to the limit, preparing Lecture 7's Bolzano–Weierstrass theorem.

The thread is becoming clear: definition (this lecture) → sufficient condition (monotone plus bounded) → intrinsic criterion (Cauchy) → compactness (Bolzano–Weierstrass).